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For n not equal to -1, it is 1/(n+1)*xn+1 while for n = -1, it is ln(|x|), the logarithm to base e.

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7y ago

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7y ago

That is:(x to the power (n+1)) / (n + 1) + C

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6y ago

It is (x^(n+1))/(n+1)

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Q: What is the integral of x to the power n?
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Related questions

Integral of e to the power of -x?

integral of e to the power -x is -e to the power -x


What is the integral of x to the power of n with respect to x?

∫ xn dx = xn+1/(n+1) + C (n ≠ -1) C is the constant of integration.


What is the integral of a function of x raised to the power of n multiplied by the derivative with respect to x of that same function of x with respect to x?

∫ f(x)nf'(x) dx = f(x)n + 1/(n + 1) + C n ≠ -1 C is the constant of integration.


What is the answer of 1 divide by x square?

What do you mean? As this is a calculus question, I presume that you are asking for a derivative or integral The derivative of any function of the form ƒ(x) = a * x ^ n is ƒ'(x) = a * n * x ^ (n-1) The integral of any function of the form ∫ a*x ^ n is a / (n+1) * x ^ (n+1) + C Your function that you gave is 1 / x^(2) which is equal to: x^(-2) Thus the derivative is: -2 * x^(-3) And the integral is: -x^(-1) + C


What would be the integral of x where x has power 2?

If x has the power 2 then you want the integral of x2, I think. When you integrate this you get : x3/3 , plus a constant.


How do you find the integrals of x?

You add one to the power, and then divide by the power that it has so you would have: Integral of x = (x^2)/2 Integral of x^2 = (x^3)/3 Etc.


What is the integral of the derivative with respect to x of the function f multiplied by the quantity a times f plus b raised to the power of n with respect to x?

∫ f'(x)(af(x) + b)n dx = (af(x) + b)n + 1/[a(n + 1)] + C C is the constant of integration.


Integral of e to the power of x?

The antiderivative, or indefinite integral, of ex, is ex + C.


What is the integral of f divided by the quantity f plus g raised to the power of n with respect to x where f and g are functions of x?

∫ f(x)/[f(x) + g(x)]n dx = ∫ 1/[f(x) + g(x)]n - 1 dx - ∫ g(x)/[f(x) + g(x)]n dx


What is the double integral of 2x?

The integral of a single term of a polynomial, in the form of AxN is (A/N+1) x (N+1). The first integral of 2x is x2 + C. The second integral of 2x is the first integral of x2 + C, which is 1/3x3 + Cx + C.


What is the integral of 6x?

Add 1 to the coefficient and divide by that term. int[f(X) by the power rule is this.......X(n + 1)/nint(6X)= 6/2(X2)= 3X2 + C--------------


What is the integral of x raised to power x?

Using information from the Wolframalpha site. It seems that this integral can't be expressed as a finite amount of standard functions; you can go to the Wolfram Alpha site, and type "integral x^x", to get a series expansion if you are interested.